leetcode -- Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

 [易错点]

1.递归终止条件不完整，当输入是：

[2], 1 即可能递归会一直进行下去导致栈溢出

2. line 29 的循环初始条件：
开始的写法是i每次递归时都是从0开始，在输入是[1,2], 3， 产生输出：[[1,1,1],[1,2],[2,1]]
正确的做法应该是从depth开始

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        int len = candidates.length, depth = 0;
        if(len == 0){
            return result;
        }
        ArrayList<Integer> output = new ArrayList<Integer>();
        int sum = 0;
        Arrays.sort(candidates);
        generate(result, output, sum, depth, len, target, candidates);
        return result;
    }
    
    public void generate(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> output, int sum,
            int depth, int len, int target, int[] candidates){
19             if(sum > target){
20                 return;
21             }
            if(sum == target){
                ArrayList<Integer> tmp = new ArrayList<Integer>();
                tmp.addAll(output);
                result.add(tmp);
                return;
            }
            
            for(int i = depth; i < len; i++){
                sum += candidates[i];
                output.add(candidates[i]);
                generate(result, output, sum, i, len, target, candidates);
                sum -= output.get(output.size() - 1);
                output.remove(output.size() - 1);
            }
        }
}